normal groups...

Hello... I'm having difficulties solving the next problem:
Let G be a group, and N<G is a normal sub-group of G.
I'm given that |N|=n, and [G:N]=m.
a) prove that to every g in G, g^m is an elemnt in N.
I've proved that.
b) Now say that gcd(m,n) = 1. Prove that N is the only sub-group of G with order n.

So, I'm having trouble with "b"... I don't seem to get anywhere.

And another question: I'm not given that G is finite, but is it not a consequence of "n" and "m" being finite? G is spanned by a finite number of cosets with a finite number of elements - I guess that means G cannot be infinite....

Let G be a group, and N<G is a normal sub-group of G.
I'm given that |N|=n, and [G:N]=m.

b) Now say that gcd(m,n) = 1. Prove that N is the only sub-group of G with order n.

You are correct. If and then is finite, and furthermore, .

Let be a subgroup of order . Since is normal subgroup of we can form a factor group . The order of this factor group is . Let . And consider the coset . Lagrange's theorem tells us that divides . But since and it means (again by Lagrange's theorem). Hence, . Thus, divides (by the property of orders). But since it means is forced to be . Thus, and so . This means . And this completes the proof (that i.e. it is unique) because if it were the case that then it would mean which is impossible.

Here is a related question (a converse statement if you wish to think of it that way). Let G be any group (even infinite) and it has a unique subgroup of index n prove that this subgroup is normal. (Nothing complicated if you approach it correctly).